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3 years ago

There are 2.2 pounds in a kilogram. if a boys mass is 40kg, what is his height in pounds?

2 answers:

4 0

There are NOT 2.2 pounds in a kilogram.
Any teacher who makes such a statement is misleading you, and
putting things into your head that will be tough to get rid of later.

2.2 pounds is the approximate WEIGHT of 1 kilogram of MASS
on Earth. The SAME KILOGRAM of mass weighs about 5.8 ounces
on the Moon, and about 11.8 ounces on Mars.

If a boy's mass is 40 kg, then he weighs about 29.2 pounds on Mars,
about 14.7 pounds on the Moon, and about 89 pounds on Earth.

"Height" doesn't belong anywhere in this question.

sasho [114]3 years ago

3 0

The simplest way to do this is to set up equivalent fractions, like this-

$\frac{1}{2.2}$ = $\frac{40}{x}$

Solve for x by using cross multiplication.

40*2.2= 88
1*x=88
x=88

Therefore, the boy weighs 88lbs.

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Kieran takes off from rest down a 50 m high, 10° slope on his jet-powered skis. The skis have a thrust of 280 N parallel to the

xz_007 [3.2K]

Answer:0.46

Explanation:

Given

Initial height $h=50 m$

inclination $\theta =10^{\circ}$

Thrust $=280 N$

combined mass of kieran and skis $m=50 kg$

Speed at the bottom $v=40 m/s$

From Work Energy Theorem

Work done by all the force is equal to change in kinetic Energy

$W_{gravity}+W_{friction}+W_{thrust}=\frac{1}{2}mv^2-0$ ------------1

distance traveled along the slope $x=\frac{50}{\sin 10}=287.93 m$

$W_{gravity}=mgh=50\times 9.8\times 50=24500 J$

$W_{thrust}=F\times x=280\times 287.93=80,620.4 J$

$W_{friction}=-\mu mg\cos 10$

substitute in 1

$24,500+80,620.4+W_{friction}=\frac{1}{2}\times 50\times 1600$

$W_{friction}=40,000-24,500-80,620.4$

$-\mu \cdot 50\times 9.8\times 287.93=-65,120.4$

$\mu =\frac{65,120.4}{141,085.7}=0.46$

8 0

3 years ago

do batteries produce or make energy

melamori03 [73]

Batteries Produce energy

3 0

3 years ago

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A steady current I flows through a wire of radius a. The current density in the wire varies with r as J = kr, where k is a const

grin007 [14]

Answer:

Explanation:

we can consider an element of radius r < a and thickness dr. and Area of this element is

$dA=2\pi r dr$

since current density is given

$J=kr$

then , current through this element will be,

$di_{thru}=JdA=(kr)(2\pi\,r\,dr)=2\pi\,kr^2\,dr$

integrating on both sides between the appropriate limits,

$\int_0^Idi_{thru}=\int_0^a2\pi\,kr^2\,dr
\\\\
I=\frac{2\pi\,ka^3}{3} -------------------------------(1)$

Magnetic field can be found by using Ampere's law

$\oint{\vec{B}\cdot\,d\vec{l}}=\mu_0\,i_{enc}$

for points inside the wire ( r<a)

now, consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius r.

by applying the Ampere's law, we can write

$\oint{\vec{B}_{in}\cdot\,d\vec{l}}=\mu_0\,i_{enc}
$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop.

Hence,

$B_{in}\times2\pi\,l=\mu_0\int_0^r(kr)(2\pi\,r\,dr)=
\\\\2\pi\,B_{in} l=2\pi\mu_0k \frac{r^3}{3}
\\\\B_{in}=\frac{\mu_0kl^2}{3}
$

now using equation 1, putting the value of k,

$B_{in} = \frac{\mu_{0} l^2 }{3 } \,\,\, \frac{3I}{2 \pi a^3}
\\\\B_{in} = \frac{ \mu_{0} I l^2}{2 \pi a^3}
$

now, for points outside the wire ( r>a)

consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius l.

applying the Ampere's law

$\oint{\vec{B}_{out}\cdot\,d\vec{l}}=\mu_0\,i_{enc}
$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop. Hence

$B_{out}\times2\pi\,r=\mu_0\int_0^a(kr)(2\pi\,r\,dr)
\\\\2\pi\,B_{out}r=2\pi\mu_0k\frac{a^3}{3}
\\\\B_{out}=\frac{\mu_0ka^3}{3r}
$

again using,equaiton 1,

$B_{out}= \mu_0 \frac{a^3}{3r} \times \frac{3 I}{2 \pi a^3}
\\\\B_{out} = \frac{ \mu_{0} I}{2 \pi r}$

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3 years ago

What is the ratio of escape speed from earth to circular orbital speed? ignore air resistance.

klio [65]

About 40 000 km/h
Here you go:

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3 years ago

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A LED light source contains a 0.5-Watts GaAs (Eg =1.43 eV) LED. Assuming that 0.12% of the electric energy is converted to emiss

Ray Of Light [21]

Answer:

Explanation:

energy emitted by source per second = .5 J

Eg = 1.43 eV .

Energy converted into radiation = .5 x .12 = .06 J

energy of one photon = 1.43 eV

= 1.43 x 1.6 x 10⁻¹⁹ J

= 2.288 x 10⁻¹⁹ J .

no of photons generated = .06 / 2.288 x 10⁻¹⁹

= 2.6223 x 10¹⁷

wavelength of photon λ = 1275 / 1.43 nm

= 891.6 nm .

momentum of photon = h / λ ; h is plank's constant

= 6.6 x 10⁻³⁴ / 891.6 x 10⁻⁹

= .0074 x 10⁻²⁵ J.s

Total momentum of all the photons generated

= .0074 x 10⁻²⁵ x 2.6223 x 10¹⁷

= .0194 x 10⁻⁸ Js

b ) spectral width in terms of wavelength = 30 nm

frequency width = ?

n = c / λ , n is frequency , c is velocity of light and λ is wavelength

differentiating both sides

dn = c x dλ / λ²

given dλ = 30 nm

λ = 891.6 nm

dn = 3 x 10⁸ x 30 x 10⁻⁹ / ( 891.6 x 10⁻⁹ )²

= 11.3 x 10¹² Hz .

c )

10 nW = 10 x 10⁻⁹ W

= 10⁻⁸ W .

energy of 50 dB

50 dB = 5 B

I / I₀ = 10⁵ ; decibel scale is logarithmic , I is energy of sound having dB = 50 and I₀ = 10⁻¹² W /s

I = I₀ x 10⁵

= 10⁻¹² x 10⁵

= 10⁻⁷ W

= 10 x 10⁻⁸ W

power required

= 10⁻⁸ + 10 x 10⁻⁸ W

= 11 x 10⁻⁸ W.

5 0

3 years ago